Determine whether the sequences that follow are bounded, monotonic and/or eventually monotonic. Determine whether each sequence converges or diverges. If the sequence converges, find its limit.

$\left\{\frac{2k^2+1}{3k^2-1}\right\}$

Consider the given question,

$\left\{\frac{2k^2+1}{3k^2-1}\right\}$

In the sequence $\left\{a_k\right\}=\left\{\frac{2k^2+1}{3k^2-1}\right\}$then the general term is given below,

$a_k=\frac{2k^2+1}{3k^2-1}$.

The term $a_{k+1}-a_k$gives, by substitution,

localid="1650647702912" $$\begin{gathered} a_{k+1}-a_k=\frac{2{\left(k+1\right)}^2+1}{3{\left(k+1\right)}^2-1}-\frac{2k^2+1}{3k^2-1} \\ =\frac{2k^2+4k+3}{3k^2+6k+2}-\frac{2k^2+1}{3k^2-1} \\ =\frac{\left(2k^2+4k+3\right)\left(3k^2-1\right)-\left(2k^2+1\right)\left(3k^2+6k+2\right)}{\left(3k^2+6k+2\right)\left(3k^2-1\right)} \\ =-\frac{10k+5}{\left(3k^2+6k+2\right)\left(3k^2-1\right)}......\left(i\right) \end{gathered}$$

Equation (i) less than $0$as $k>0$. Thus,$a_{k+1}-a_k$.

The sequence is strictly decreasing. Hence, it is monotonic.

The sequence is bounded below because $0<a_k$as $k>1$.

As the index k increases, the term approaches role="math" localid="1650647934211" $$\begin{gathered} a_k=\frac{k^2-2}{k^2+2k+2} \\ =0.67 \end{gathered}$$.

As $a_k<0.6$. Then, the increasing sequence has an upper bound and is $0.67$.

Thus, $0<a_k<0.67$.

The given sequence has lower and upper bounds. Therefore, the sequence is bounded.

The monotonic decreasing sequence with lower bound is convergent.

The monotonic decreasing the given sequence is bounded below and is convergent.

Therefore, the sequence is convergent.

To find the limit of the given sequence,

$$\begin{gathered} \underset{k\to \infty }{\lim }{a}_k=\underset{k\to \infty }{\lim }\left(\frac{2k^2+1}{3k^2-1}\right) \\ =\underset{k\to \infty }{\lim }\left(\frac{2+{\displaystyle \frac{1}{k^2}}}{3-\frac{1}{k^2}}\right) \\ =\frac{2}{3} \end{gathered}$$

Thus, the limit of the given sequence is$\frac{2}{3}$.